$\sum\limits_{n=1}^{\infty } \dfrac{(x+4)^n}{n\cdot4^n} $ What is the interval of convergence of the series? Choose 1 answer: Choose 1 answer: (Choice A) A $-6 \le x<-2$ (Choice B) B $-8 \le x<0$ (Choice C) C $-6 < x<-2$ (Choice D) D $-8 < x<0$
We use the ratio test. For $x\neq-4$, let $a_n= \dfrac{(x+4)^n}{n\cdot4^n} $. $\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|= \left| \dfrac{x+4}{4}\right|$ The series converges when $\left| \dfrac{x+4}{4}\right| <1$, which is when $-8<x<0$. Now let's check the endpoints, $x=-8$ and $x=0$. Letting $x=-8$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(-8+4)^n}{n\cdot4^n} &= \sum\limits_{n=1}^{\infty } \dfrac{(-4)^n}{n\cdot4^n} \\\\ &= \sum\limits_{n=1}^{\infty } \dfrac{(-1)^n(4)^n}{n\cdot4^n} \\\\ &= \sum\limits_{n=1}^{\infty } \dfrac{(-1)^n}{n} \end{aligned}$ This is the alternating harmonic series, which is known to converge. Letting $x=0$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(0+4)^n}{n\cdot4^n} &= \sum\limits_{n=1}^{\infty } \dfrac{4^n}{n\cdot4^n} \\\\ &= \sum\limits_{n=1}^{\infty } \dfrac{1}{n} \end{aligned}$ This is the harmonic series, which is known to diverge. In conclusion, the interval of convergence is $-8 \le x<0$.